Indefinite Integrals Trig Quiz - Powers, Fractions & Trig

1) Evaluate: ∫ (x⁷ - 4x³ + x) dx

(1/8)x⁸- x⁴ + (½)x² + C

7x⁶ - 12x² + 1 + C

(1/8)x⁸- (1/4)x⁴ + (½)x² + C

(1/8)x⁸- x⁴ + x² + C

Apply the power rule to each term. For x⁷: ∫x⁷ dx = (1/8)x^8. For -4x³: ∫-4x³ dx = -4*(1/4)x⁴ = -x⁴. For x: ∫x dx = (½)x². Combine and add C: (1/8)x⁸- x⁴ + (½)x² + C.

Explanation

Apply the power rule to each term. For x⁷: ∫x⁷ dx = (1/8)x^8. For -4x³: ∫-4x³ dx = -4*(1/4)x⁴ = -x⁴. For x: ∫x dx = (½)x². Combine and add C: (1/8)x⁸- x⁴ + (½)x² + C.

2) Evaluate: ∫ (3 sin(x) - 4 cos(x)) dx

-3 cos(x) - 4 sin(x) + C

3 cos(x) + 4 sin(x) + C

-3 cos(x) + 4 sin(x) + C

3 cos(x) - 4 sin(x) + C

We know ∫ sin(x) dx = -cos(x) and ∫ cos(x) dx = sin(x). So, ∫3 sin(x) dx = 3*(-cos(x)) = -3 cos(x). ∫ -4 cos(x) dx = -4 * sin(x) = -4 sin(x). Combine: -3 cos(x) - 4 sin(x) + C.

Explanation

We know ∫ sin(x) dx = -cos(x) and ∫ cos(x) dx = sin(x). So, ∫3 sin(x) dx = 3*(-cos(x)) = -3 cos(x). ∫ -4 cos(x) dx = -4 * sin(x) = -4 sin(x). Combine: -3 cos(x) - 4 sin(x) + C.

3) The power rule for integration, ∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C, is valid for:

All real numbers n.

All real numbers n except n = -1.

Only positive integers n.

Only rational numbers n.

The power rule works for any real number n ≠ -1. When n = -1, the integrand is 1/x, and the antiderivative is ln|x|, not a power function. The formula would involve division by zero if n = -1. So the exception is n = -1.

Explanation

The power rule works for any real number n ≠ -1. When n = -1, the integrand is 1/x, and the antiderivative is ln|x|, not a power function. The formula would involve division by zero if n = -1. So the exception is n = -1.

4) Evaluate: ∫ (5 - 2x + 3x²) dx

5x - x² + x³ + C

5x - 2x² + 3x³ + C

5x - x² + 3x³ + C

5x - 2x² + x³ + C

Integrate term by term. ∫5 dx = 5x. ∫ -2x dx = -2*(½)x² = -x². ∫ 3x² dx = 3*(⅓)x³ = x³. So the result is 5x - x² + x³ + C.

Explanation

Integrate term by term. ∫5 dx = 5x. ∫ -2x dx = -2*(½)x² = -x². ∫ 3x² dx = 3*(⅓)x³ = x³. So the result is 5x - x² + x³ + C.

5) What is the antiderivative of f(x) = 0?

0

1

C (any constant)

X

The derivative of any constant function is 0. Therefore, the antiderivative of 0 is any constant. We write ∫ 0 dx = C, where C is an arbitrary constant. Option A, 0, is one specific antiderivative, but the general antiderivative includes all constants.

Explanation

The derivative of any constant function is 0. Therefore, the antiderivative of 0 is any constant. We write ∫ 0 dx = C, where C is an arbitrary constant. Option A, 0, is one specific antiderivative, but the general antiderivative includes all constants.

6) Which operation is the inverse of differentiation?

Integration

Multiplication

Composition

Logarithm

Differentiation finds the derivative of a function. Integration, specifically finding the indefinite integral, finds a function whose derivative is given. This reverses the process of differentiation. Therefore, integration is the inverse operation of differentiation.

Explanation

Differentiation finds the derivative of a function. Integration, specifically finding the indefinite integral, finds a function whose derivative is given. This reverses the process of differentiation. Therefore, integration is the inverse operation of differentiation.

7) Evaluate: ∫ (½) eˣ dx

(½) eˣ + C

2eˣ + C

(½) e^(x+1) + C

Eˣ + C

The integral of eˣ is eˣ. For a constant multiple, we have ∫ k * f(x) dx = k * ∫ f(x) dx. Here, k = 1/2. So ∫ (½) eˣ dx = (½) ∫ eˣ dx = (½) eˣ + C.

Explanation

The integral of eˣ is eˣ. For a constant multiple, we have ∫ k * f(x) dx = k * ∫ f(x) dx. Here, k = 1/2. So ∫ (½) eˣ dx = (½) ∫ eˣ dx = (½) eˣ + C.

8) Find ∫ (5x⁴ + 2/x²) dx.

X⁵ - 2/x + C

20x³ - 4/x³ + C

X⁵ + 2/x + C

X⁵ - 2/x³ + C

Rewrite the integrand: 5x⁴ + 2/x² = 5x⁴ + 2x^(-2). Now integrate term by term. ∫5x⁴ dx = 5*(1/5)x⁵ = x⁵. ∫2x^(-2) dx = 2 * [x^(-2+1)/(-2+1)] = 2 * [x^(-1)/(-1)] = -2x^(-1) = -2/x. Add C: x⁵ - 2/x + C.

Explanation

Rewrite the integrand: 5x⁴ + 2/x² = 5x⁴ + 2x^(-2). Now integrate term by term. ∫5x⁴ dx = 5*(1/5)x⁵ = x⁵. ∫2x^(-2) dx = 2 * [x^(-2+1)/(-2+1)] = 2 * [x^(-1)/(-1)] = -2x^(-1) = -2/x. Add C: x⁵ - 2/x + C.

9) The indefinite integral ∫ f(x) dx represents:

The area under the curve of f(x) from a to b.

The slope of the tangent line to f(x).

The general antiderivative of f(x).

The derivative of f(x).

By definition, the indefinite integral ∫ f(x) dx is the collection of all antiderivatives of f(x). It is expressed as F(x) + C, where F'(x) = f(x). Option A describes a definite integral. Option B describes the derivative. Option D is the operation opposite to integration.

Explanation

By definition, the indefinite integral ∫ f(x) dx is the collection of all antiderivatives of f(x). It is expressed as F(x) + C, where F'(x) = f(x). Option A describes a definite integral. Option B describes the derivative. Option D is the operation opposite to integration.

10) If F'(x) = f(x), then f(x) is called the ____________ of F(x).

Derivative

Antiderivative

Integral

Constant

This is the definition of derivative. If F'(x) = f(x), then f(x) is the derivative of F(x). The antiderivative of f(x) would be F(x). The integral of f(x) would be F(x) + C.

Explanation

This is the definition of derivative. If F'(x) = f(x), then f(x) is the derivative of F(x). The antiderivative of f(x) would be F(x). The integral of f(x) would be F(x) + C.

11) Evaluate: ∫ ( (x² + 1) / x ) dx, for x > 0.

(½)x² + ln|x| + C

X + 1/x + C

(½)x² + 1/x² + C

X²/2 + ln x + C

First simplify the integrand: (x² + 1)/x = x²/x + 1/x = x + 1/x. Now integrate term by term: ∫ x dx = (½)x². ∫ (1/x) dx = ln|x|. Since x > 0, this is ln(x). So the result is (½)x² + ln(x) + C.

Explanation

First simplify the integrand: (x² + 1)/x = x²/x + 1/x = x + 1/x. Now integrate term by term: ∫ x dx = (½)x². ∫ (1/x) dx = ln|x|. Since x > 0, this is ln(x). So the result is (½)x² + ln(x) + C.

12) Which of the following functions does NOT have an elementary antiderivative?

X⁵

Sin(3x)

E^(-x²)

1/(x²+1)

While x⁵, sin(3x), and 1/(x²+1) have antiderivatives expressible in terms of elementary functions (polynomials, trigonometric, and arctangent respectively), the function e^(-x²) is known to not have an antiderivative that can be written as a finite combination of elementary functions. Its integral is related to the error function, which is not elementary.

Explanation

While x⁵, sin(3x), and 1/(x²+1) have antiderivatives expressible in terms of elementary functions (polynomials, trigonometric, and arctangent respectively), the function e^(-x²) is known to not have an antiderivative that can be written as a finite combination of elementary functions. Its integral is related to the error function, which is not elementary.

13) Evaluate: ∫ 6 sec²(x) dx

6 tan(x) + C

6 sec(x) tan(x) + C

3 tan²(x) + C

6 cos²(x) + C

We recall that the derivative of tan(x) is sec²(x). Therefore, the antiderivative of sec²(x) is tan(x). For a constant multiple, ∫ 6 sec²(x) dx = 6 ∫ sec²(x) dx = 6 tan(x) + C.

Explanation

We recall that the derivative of tan(x) is sec²(x). Therefore, the antiderivative of sec²(x) is tan(x). For a constant multiple, ∫ 6 sec²(x) dx = 6 ∫ sec²(x) dx = 6 tan(x) + C.

14) Which statement provides the most precise definition of F(x) being an antiderivative of f(x)?

F(x) is the area under the curve of f(x).

F'(x) = f(x).

F'(x) = F(x).

F(x) = f(x) + C.

The definition of an antiderivative is based on reversing the process of differentiation. A function F(x) is called an antiderivative of f(x) if the derivative of F(x) is equal to f(x). Option A describes a definite integral, not an antiderivative. Option C reverses the relationship. Option D incorrectly relates the functions through addition rather than differentiation.

Explanation

The definition of an antiderivative is based on reversing the process of differentiation. A function F(x) is called an antiderivative of f(x) if the derivative of F(x) is equal to f(x). Option A describes a definite integral, not an antiderivative. Option C reverses the relationship. Option D incorrectly relates the functions through addition rather than differentiation.